3.317 \(\int \frac{(1+x^4)^{3/4}}{(2+x^4)^2} \, dx\)

Optimal. Leaf size=74 \[ \frac{\left (x^4+1\right )^{3/4} x}{8 \left (x^4+2\right )}+\frac{3 \tan ^{-1}\left (\frac{x}{\sqrt [4]{2} \sqrt [4]{x^4+1}}\right )}{16\ 2^{3/4}}+\frac{3 \tanh ^{-1}\left (\frac{x}{\sqrt [4]{2} \sqrt [4]{x^4+1}}\right )}{16\ 2^{3/4}} \]

[Out]

(x*(1 + x^4)^(3/4))/(8*(2 + x^4)) + (3*ArcTan[x/(2^(1/4)*(1 + x^4)^(1/4))])/(16*2^(3/4)) + (3*ArcTanh[x/(2^(1/
4)*(1 + x^4)^(1/4))])/(16*2^(3/4))

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Rubi [A]  time = 0.0243437, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.294, Rules used = {378, 377, 212, 206, 203} \[ \frac{\left (x^4+1\right )^{3/4} x}{8 \left (x^4+2\right )}+\frac{3 \tan ^{-1}\left (\frac{x}{\sqrt [4]{2} \sqrt [4]{x^4+1}}\right )}{16\ 2^{3/4}}+\frac{3 \tanh ^{-1}\left (\frac{x}{\sqrt [4]{2} \sqrt [4]{x^4+1}}\right )}{16\ 2^{3/4}} \]

Antiderivative was successfully verified.

[In]

Int[(1 + x^4)^(3/4)/(2 + x^4)^2,x]

[Out]

(x*(1 + x^4)^(3/4))/(8*(2 + x^4)) + (3*ArcTan[x/(2^(1/4)*(1 + x^4)^(1/4))])/(16*2^(3/4)) + (3*ArcTanh[x/(2^(1/
4)*(1 + x^4)^(1/4))])/(16*2^(3/4))

Rule 378

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1)*(c
+ d*x^n)^q)/(a*n*(p + 1)), x] - Dist[(c*q)/(a*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[n*(p + q + 1) + 1, 0] && GtQ[q, 0] && NeQ[p, -1]

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 212

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&
!GtQ[a/b, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\left (1+x^4\right )^{3/4}}{\left (2+x^4\right )^2} \, dx &=\frac{x \left (1+x^4\right )^{3/4}}{8 \left (2+x^4\right )}+\frac{3}{8} \int \frac{1}{\sqrt [4]{1+x^4} \left (2+x^4\right )} \, dx\\ &=\frac{x \left (1+x^4\right )^{3/4}}{8 \left (2+x^4\right )}+\frac{3}{8} \operatorname{Subst}\left (\int \frac{1}{2-x^4} \, dx,x,\frac{x}{\sqrt [4]{1+x^4}}\right )\\ &=\frac{x \left (1+x^4\right )^{3/4}}{8 \left (2+x^4\right )}+\frac{3 \operatorname{Subst}\left (\int \frac{1}{\sqrt{2}-x^2} \, dx,x,\frac{x}{\sqrt [4]{1+x^4}}\right )}{16 \sqrt{2}}+\frac{3 \operatorname{Subst}\left (\int \frac{1}{\sqrt{2}+x^2} \, dx,x,\frac{x}{\sqrt [4]{1+x^4}}\right )}{16 \sqrt{2}}\\ &=\frac{x \left (1+x^4\right )^{3/4}}{8 \left (2+x^4\right )}+\frac{3 \tan ^{-1}\left (\frac{x}{\sqrt [4]{2} \sqrt [4]{1+x^4}}\right )}{16\ 2^{3/4}}+\frac{3 \tanh ^{-1}\left (\frac{x}{\sqrt [4]{2} \sqrt [4]{1+x^4}}\right )}{16\ 2^{3/4}}\\ \end{align*}

Mathematica [C]  time = 0.0097326, size = 41, normalized size = 0.55 \[ \frac{x \, _2F_1\left (-\frac{3}{4},\frac{1}{4};\frac{5}{4};-\frac{x^4}{x^4+2}\right )}{2\ 2^{3/4} \sqrt [4]{x^4+2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(1 + x^4)^(3/4)/(2 + x^4)^2,x]

[Out]

(x*Hypergeometric2F1[-3/4, 1/4, 5/4, -(x^4/(2 + x^4))])/(2*2^(3/4)*(2 + x^4)^(1/4))

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Maple [F]  time = 0.033, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{ \left ({x}^{4}+2 \right ) ^{2}} \left ({x}^{4}+1 \right ) ^{{\frac{3}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4+1)^(3/4)/(x^4+2)^2,x)

[Out]

int((x^4+1)^(3/4)/(x^4+2)^2,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (x^{4} + 1\right )}^{\frac{3}{4}}}{{\left (x^{4} + 2\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4+1)^(3/4)/(x^4+2)^2,x, algorithm="maxima")

[Out]

integrate((x^4 + 1)^(3/4)/(x^4 + 2)^2, x)

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Fricas [B]  time = 52.7589, size = 668, normalized size = 9.03 \begin{align*} -\frac{12 \cdot 8^{\frac{3}{4}}{\left (x^{4} + 2\right )} \arctan \left (-\frac{8^{\frac{3}{4}}{\left (x^{4} + 1\right )}^{\frac{1}{4}} x^{3} + 4 \cdot 8^{\frac{1}{4}}{\left (x^{4} + 1\right )}^{\frac{3}{4}} x - 2^{\frac{1}{4}}{\left (8^{\frac{3}{4}} \sqrt{x^{4} + 1} x^{2} + 8^{\frac{1}{4}}{\left (3 \, x^{4} + 2\right )}\right )}}{2 \,{\left (x^{4} + 2\right )}}\right ) - 3 \cdot 8^{\frac{3}{4}}{\left (x^{4} + 2\right )} \log \left (\frac{8 \, \sqrt{2}{\left (x^{4} + 1\right )}^{\frac{1}{4}} x^{3} + 8 \cdot 8^{\frac{1}{4}} \sqrt{x^{4} + 1} x^{2} + 8^{\frac{3}{4}}{\left (3 \, x^{4} + 2\right )} + 16 \,{\left (x^{4} + 1\right )}^{\frac{3}{4}} x}{x^{4} + 2}\right ) + 3 \cdot 8^{\frac{3}{4}}{\left (x^{4} + 2\right )} \log \left (\frac{8 \, \sqrt{2}{\left (x^{4} + 1\right )}^{\frac{1}{4}} x^{3} - 8 \cdot 8^{\frac{1}{4}} \sqrt{x^{4} + 1} x^{2} - 8^{\frac{3}{4}}{\left (3 \, x^{4} + 2\right )} + 16 \,{\left (x^{4} + 1\right )}^{\frac{3}{4}} x}{x^{4} + 2}\right ) - 64 \,{\left (x^{4} + 1\right )}^{\frac{3}{4}} x}{512 \,{\left (x^{4} + 2\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4+1)^(3/4)/(x^4+2)^2,x, algorithm="fricas")

[Out]

-1/512*(12*8^(3/4)*(x^4 + 2)*arctan(-1/2*(8^(3/4)*(x^4 + 1)^(1/4)*x^3 + 4*8^(1/4)*(x^4 + 1)^(3/4)*x - 2^(1/4)*
(8^(3/4)*sqrt(x^4 + 1)*x^2 + 8^(1/4)*(3*x^4 + 2)))/(x^4 + 2)) - 3*8^(3/4)*(x^4 + 2)*log((8*sqrt(2)*(x^4 + 1)^(
1/4)*x^3 + 8*8^(1/4)*sqrt(x^4 + 1)*x^2 + 8^(3/4)*(3*x^4 + 2) + 16*(x^4 + 1)^(3/4)*x)/(x^4 + 2)) + 3*8^(3/4)*(x
^4 + 2)*log((8*sqrt(2)*(x^4 + 1)^(1/4)*x^3 - 8*8^(1/4)*sqrt(x^4 + 1)*x^2 - 8^(3/4)*(3*x^4 + 2) + 16*(x^4 + 1)^
(3/4)*x)/(x^4 + 2)) - 64*(x^4 + 1)^(3/4)*x)/(x^4 + 2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (x^{4} + 1\right )^{\frac{3}{4}}}{\left (x^{4} + 2\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**4+1)**(3/4)/(x**4+2)**2,x)

[Out]

Integral((x**4 + 1)**(3/4)/(x**4 + 2)**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (x^{4} + 1\right )}^{\frac{3}{4}}}{{\left (x^{4} + 2\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4+1)^(3/4)/(x^4+2)^2,x, algorithm="giac")

[Out]

integrate((x^4 + 1)^(3/4)/(x^4 + 2)^2, x)